not all code path return a value

Question

hi,
when i use the codings i have an error like not all code path return a value.how can i solve this
public string try{        if          {                 companymailid = mailname;                 Session[ "cmailid" ] = companymailid;                 Server.Transfer(           }              {                   companymailid = "" ;               }
} else      return companymailid;} "mail.aspx" ); (mailname != null ) companymailid; string mailid( string mailname){

Manikavelu Velayutham · Accepted Answer

In your code its seems to be Only the Try block returns the value and not the Catch Block, thats the cause of the error.
Try like this
public string test()
{
try         {                  return "" ;
}
catch         {                  return "" ;
} }

Suthish Nair · Answer

just use Response.Redirect("mail.aspx");

Jaganathan Bantheswaran · Answer

Hi,
Try this.
try
{
if(somecondition)
{
return something; //optional
}
else
{
return something; // optional
}
}
catch(Exception)
{
}
return something; //else error will raise.
the first two return statements are not considering in this case. because of its inside the if block.

karthi vimala · Answer

hi,
ya,i specify false parameter in server.transfer("mail.aspx",false) and also for mail.aspx page i set enabledviewstate is false now also the same error will occur.
i didn't know why the error will occur.can anyone explain me why it will occur.

Suthish Nair · Answer

Instead of Transfer use Server.Execute or Response.Redirect("123.aspx",false)

Krishna Garad · Answer

Can you fix it what error is occuring. I mean you are saying in post that "Not all code path returns value" it is one of the compile time error but in your attachment you are showing error which generally occurs runtime.

karthi vimala · Answer

hi,
i correct the previous error,by using session i pass the value to mail.aspx page when i execute this th error
Error executing child request for mail.aspx.
will occur,how can i solve this
here i attached my codings

Vulpes · Answer

Unfortunately, your code has come out as a muddle when you pasted it in and is therefore difficult to follow.
However, it looks to me as though there is no return statement in the 'if' block but there is a return statement in the 'else' block. Correcting the error may therefore be as simple as placing the line: return companymailid; at the end of the 'if' block.